PLEASE HELP ME!! Q 1) It’s been a long day and you are working late in the lab.
ID: 217336 • Letter: P
Question
PLEASE HELP ME!!
Q 1) It’s been a long day and you are working late in the lab. You have set up a bacterial culture for a critical experiment and you need to harvest them when they reach a cell density of exactly 6 x 108 cells/ml. Unfortunately, you used a very old stationary culture to inoculate your culture and so the cells spent a lot of time in lag phase before they entered into exponential growth. Consequently, it’s now 8:30 pm, you haven’t had dinner and you are very hungry. It suddenly occurs to you that there’s a sandwich place that’s open till 9 pm and you can get there on your bike before closing time and make it back to the lab by 9:30 pm.
You really want that sandwich but if your cells reach a density of > 6 x 108 cells/ml before you get back they cannot be used for the bioassay and you will not only have to start all over again tomorrow but will also have to endure snarky comments from your lab supervisor who will point out that you should have planned things better.
When you measured the cell density at 8:30, it was 2 x 108 cells/ml. The bacteria you are growing have a generation time of 60 min:
Q 2) You are doing a viable cell count on a bacterial culture and you perform ten-fold serial dilutions as we did in class (Exercise 11). You spread 0.2 ml of each dilution on BHI plates and, after incubating at 370 overnight, obtain the results shown below. How many viable cells per milliliter are there in the undiluted bacterial culture? Show your work.
Dilution: 10-2, 10-3, 10-4, 10-5, 10-6, 10-7, 10-8
CFU: TNTC, TNTC, TNTC, 579, 61, 8, 1
Explanation / Answer
Please find the answers below:
Answer 1: According to the formula for calculating the number of microbes available in the culture after a given point of time, we can have:
Nt = N0 * egt where, Nt = number of cells after time t, t = time (in hours), g = growth rate (in hours), N0 = initial number of cells.
Thus, we can solve the equation for obtaining t in this case as:
t = ln (Nt/N0)/g
thus, t = ln( 6*108/2*108)/1
thus, t = ln3 or t = 1.098 hours or nearly 66 minutes.
Thus, the experimenter can return back at 09:30 exactly and the cell count would reach optimum by the time.
Answer 2: The colony forming unit or CFU can be counted for a given dilution only when the number of colonies on the plate lies between 30 to 300. The number of colonies below or above this range are not considered for counting CFU.
Thus, under the present case, the plating volume is 0.2 ml and the viable cell count of 61 was observed at a dilution of 10-6, thus we can use the formula for CFU as:
CFU/ml = Number of colonies appearing / (volume plated * dilution factor)
CFU/ml = 61/ (0.2 * 10-6)
Thus, CFU/ml = 3.05 *108 in the original sample.
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